Weren't you using similar reasoning to show that angle BMO = angle ANO?

I used different reasoning. My reasoning for those two angles being congruent: The line segments AO and BO both extend from the equilateral triangle's angles to its center, thus are congruent, which makes congruent the angles BMO and ANO. That gave me two congruent angles and a non-included congruent side on each triangle, so AAS congruency.

Quote:

Originally Posted by skipjack

Opposite angles of the cyclic quadrilateral AMON are supplementary, so $\small{\angle\hspace{1px}}$MON = 120$^\circ\!$.

Thank you for this explanation. I "see" it now. Happy Holidays and Happy New Year!